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Are turn rate indicators calibrated to a certain speed? If so, what speed?
Are turn rate indicators calibrated to a certain speed? If they are, what speed are they calibrated to, and why aren't all the turn rate indicators calibrated for the same speed?

If they are calibrated to a certain speed, that's news to me. I'm not an avionics expert, but slip indicators don't have units that I've seen, and turn indicators have a mark on either side that indicate a 180 degree per minute turn; otherwise there are no units or measure points for that either.

Turn coordinators come in two flavors. The standard in GA aircraft are calibrated to a 2 minute 360° turn when the wings of the airplane in the indicator line up with the markings.  They also make 4 minute turn coordinators. The angle of bank required for a 2 minute or 4 minute turn varies with the aircraft speed—the higher the speed the greater the angle of bank. The turn and bank indicator works on the same principle, but doesn’t indicate the rate of turn when the turn first starts. I’ve never heard them called turn and slip indicators, but that makes more sense.

@Adder When you asked this question, did you mean airspeed or speed of the gyro's rotation?

Related -- https://aviation.stackexchange.com/q/65639/34686

They are not calibrated for a certain speed.
If by "calibrated", you mean they deflect together / to the same side, then you're using the instruments wrong. The two instruments have are not linked to each other, they operate independently and they operate on different principles. They are located next to each other in the cockpit out of convenience.
The "slip indicator" is a ball in a U-shape tube. It tells you whether gravity (or experienced acceleration) is pointed down, or to the side. Period. It is as simple as that.
The "turn coordinator" (little white plane / white needle) indicates your yaw rate. It operates by using a spinning gyro; the precession effect of yawing causes the gyro assembly to roll.

The roll is restricted by a spring, which is calibrated to indicate standard rate turn when the yaw rate is 3 degrees per second. The turn coordinator works at any airspeed (part of the preflight check is to turn the plane while taxing and verify the turn coordinator works).

Well, in some sense turn indicators (slip not) are calibrated for speed—the higher the speed, the lower rate of turn you get from the same bank angle, so faster aircraft have slower standard turn (4 minutes rather than 2). Jets don't use turn coordinators much any more anyway and use fixed bank angle instead.

Suggestion for improvement-- unlike what you suggested in the third paragraph, the term "turn coordinator" is generally not used for the type of instrument with the needle display.  There's a specific difference between the internal workings of a turn rate indicator (needle type of display) and "turn coordinator" (airplane type of display).

@Jan Hudec, What you said about the physics of airspeed affecting turn rate is true, but don't confuse this with calibration of the instrument.

They are calibrated for a certain true airspeed. Imagine that you are (somehow) in a 90 degree level banked turn pulling 1g. (Perhaps we turned off gravity). A turn rate indicator will read 0 since the rotation is entirely in pitch instead of yaw. But you watching your compass or an air traffic controller watching your ground track will see a very high turn rate!
This same effect, to a lesser degree, occurs at any non-zero bank angle. The turn indicator measures rotation around the vertical axis of the aircraft, but what we're really interested in is how fast our compass direction is changing. These are the same thing when the wings are level, but as soon as we bank the wings the relationship changes.
From this we might conclude that the turn indicator is dramatically wrong at indicating a standard rate turn if we change our true airspeed, and thus required bank angle, by a large amount. But there is a secondary effect. Since the gyroscope is tilted relative to the aircraft's vertical axis during a turn, it also responds to pitch to some degree. Its sensitivity depends on how tilted it is relative to the vertical axis. By judicious choice of the rotational speed of the gyroscope and/or the stiffness of the spring used it is possible to make these effects exactly cancel out for one specific bank angle. So at that one specific bank angle (and thus true airspeed) the turn indicator will indicate exactly standard rate during a standard rate turn.
Fortunately, doing so also makes the turn indicator indicate approximately correct so long as the bank angle isn't too different than the one that it was calibrated at. For instance, changing the bank angle by 10 degrees only changes the indication by about 1.5%, which is less than 3 degrees difference for a 180 degree turn. This is a small enough difference to be negligible.

As per the clear intent of the original question, this answer deals with the deflection of the needle in a turn rate indicator (or the deflection of the symbolic airplane in a turn "coordinator"), not the deflection of the slip-skid ball.  "Turn and slip indicator" or "turn and bank indicator" are names that are commonly applied to the whole package (at least in the case of the turn rate indicator with the needle rather than the turn "coordinator" with the little symbolic airplane), but it's clear that the original question is not asking about the movement of the slip-skid ball (inclinometer).
Also, this answer only deals with actual gyroscopic instruments, not their more modern counterparts.
Also, we'll assume throughout this answer that the slip-skid ball (inclinometer) is kept centered via appropriate rudder inputs by the pilot, or by the aircraft's yaw damper or other automated systems, if needed.
Also, please don't confuse the ideas explored in this answer, with the fact that various turn rate indicators designed for use in various types of aircraft with varying performance use different values for the turn rate that is achieved when the needle is at the primary index marking, and also use different values for the turn rate achieved when the needle approaches full deflection.  For example, airliners often perform half-standard-rate turns (four minutes per turn), so many turn rate indicators intended for use in such aircraft have the primary index markings set to correspond to a half-standard-rate turn.  Such instruments will be marked with the label "4 minute turn" rather than the more customary label "2 minute turn".  But that's not what this answer is about.

Are turn and slip indicators calibrated to a certain speed?

Essentially, yes.  On any given turn rate indicator, each turn rate index marking will only be completely accurate at one specific True airspeed.  On an instrument with multiple turn rate index markings, they wouldn't necessarily all need to all be calibrated to be accurate at the exact same True airspeed, though it seems logical that they might be.
Most illustrations of turn rate indicators with multiple turn rate indices appear to depict older instruments, often of British or other non-U.S. manufacture.  Many turn rate indicators bear only one index for turn rate in each direction, corresponding to either a standard-rate (two minute) or half-standard-rate (four minute) turn.  In this case the answer to the question would be "yes, on any given turn rate indicator, the turn rate index is only completely accurate at one specific True airspeed."
However, the variation in the accuracy of a turn rate indicator as True airspeed varies is small enough that it is not generally something that a pilot needs to take into account in actual practice.
But it does exist.
For example, in one source we read that:

The indicators are calibrated for rate 1, 2 and 3 turns at specified
angles of bank and TAS, and looping error is taken into account in
this calibration

Original source: "a rather elderly version of RAF publication AP1234"
Quoted in: this post on the PPRuNe aviation forum
In another source we read:

Turn rate indicators are calibrated to Rate 1, 2 and 3 turns at a
specific TAS. A rate 1 turn is 180° a minute or 3° a second. A rate 2
turn is twice that, 360° a minute or 6° a second. A rate 3 turn is
three times as much, 540° a minute or 9° a second. The angle of bank
required to achieve a given rate of turn increases with the TAS.
It is inherent in the design of the instrument that in any yaw
condition the gyro axis will tilt. and the gyro will become sensitive
to pitch rate.
If the aircraft is then rapidly pitched nose up (in a loop or recovery
from a spiral dive) this pitch input can deflect the gyro to read
maximum turn rate. This is called looping error. This pitch rate error
also affects the instrument readings in normal turns.
To compensate for this and for the difference between yaw and turn
rates the indicators are calibrated to show rates of turn correctly in
balanced turns for Rate 1, 2 and 3 turns at specific angles of bank
and TAS.
Although the indicated rate of turn will be incorrect at speeds away
from these datums the errors are not significant in normal operation.
The direction of rotation of the gyro is chosen so that when an
aircraft banks into a balanced turn the gyro precesses in the opposite
roll sense to keep its axis more or less horizontal and therefore keep
it more sensitive to turn rate.  If the gyro rotated in the opposite
direction it would only work satisfactorily at very low yaw rates and
small angles of bank.

Source: Bristol Groundschool ATPL Training Material sample, v6.1.5, page 3.13, published November 13, 2013, accessed via this on-line link, italicization and bolding added.
In other words, the calibration of the indicators to show specific turn rates is carried out at specific bank angles and specific True airspeeds.
Actually, both of these quoted statements are a little less than crystal clear, because if the turn rate is specified, then we only need to specify one additional parameter, either bank angle or True airspeed, in order to fully constrain the other parameter as well. They can't vary independently of each other, assuming that the slip-skid ball is kept centered. Still, the point is made that the calibration of the instrument for turn rate is not independent of True airspeed.
Here's a quote from an on-line discussion forum that actually cites the specific airspeed used to calibrate the turn rate indicator in one specific aircraft:

Also, if you consider that in a balanced turn you have yaw roll and
pitch, the pitch element makes the instrument over-read as the
aircraft will be at more than 1g. For this reason, the turn indicator
is calibrated to read correctly with this increased g by adjusting the
tension on the springs, but at one speed only . The instrument fitted
to the Chipmunk in RAF service was actually calibrated to 180 knots,
so I don't think this correction is too significant.

Source: this post on the PPRuNe aviation forum

and why aren't all the turn rate indicators calibrated for the same
speed?

It would make no sense that turn rate indicators intended for use in aircraft with vastly different performance would all be calibrated to have maximum accuracy at exactly the same airspeed.  Especially given the fact that the critical parameter is actually True airspeed, which varies even more widely between different aircraft types than does Indicated airspeed.
Why should the calibration of a turn rate indicator be airspeed-dependent?
The basic reason that the calibration of a turn rate indicator is airspeed-dependent is that for a given turn rate, the spin axis of the gyro wheel cannot be horizontal relative to the earth at all possible bank angles.
We can start by thinking of a turn rate indicator as a yaw rate sensor.  A turn is comprised of a pitch rotation as well as a yaw rotation.  For a given turn rate, the yaw rate is dependent on the bank angle.  For a given turn rate, the steeper the bank angle, the lower the yaw rate.  And for a given turn rate, the higher the True airspeed, the steeper bank angle.  So for a given turn rate, the yaw rate is dependent on the True airspeed.
But the story is more complicated than that, because a turn rate indicator is actually more than just a yaw rate sensor.
Let's explore this in more detail.  (This will get a bit complicated-- the reader is encouraged to make some sketches, or better yet, to make a simple model of a gyro wheel such as a pencil skewered through a round disk that he or she can hold in different attitudes in space.)
In this You Tube video entitled How Gyroscopic Turn & Slip Indicator Works, we can see what a turn rate indicator looks like inside.  At rest, the gyro's spin axis is oriented laterally (parallel to the aircraft's wingspan).  The direction of spin is clockwise as viewed from the right wingtip, so that the top of the gyro wheel is spinning away from the pilot.  The instrument's pivot axis or "gimbal axis" is oriented longitudinally-- fore and aft.
In the video, we can see the gyro's spin axis roll (bank) to a new orientation in response to yaw.  The degree of movement is rather substantial.  The rotation of the gyro wheel converts the aircraft's yaw rate to a roll torque on the gyro, forcing the gyro's spin axis to roll (bank) against the direction of the yaw, i.e. opposite to direction that the aircraft is banking in a coordinated turn.  Gears convert this movement to a motion of the pointer in the opposite direction, i.e. in the same direction that the aircraft is banking in a coordinated turn.
Note that the gyro is not acting like a "stable platform" like the artificial horizon-- purely rolling the instrument has no effect on the pointer, as it would if the gyro's spin axis were tending to stay fixed in space.  Apparently the centering springs are strong enough to overcome any such tendency.  Instead, as the aircraft yaws, the roll torque created by gyroscopic precession is acting against the centering springs to cause the gyro's spin axis to come to rest in an orientation that is "banked" relative to the instrument case.  Note also that the direction that the gyro's spin axis "banks" relative to the instrument case is also the direction that would tend to keep the gyro's spin axis more or less level (horizontal) with respect to the earth.  However, since the gyro is not acting like a stable platform, there's no reason to think that the gyro's spin axis will stay exactly horizontal with respect to the earth.  In fact, for a steady-state turn at a given turn rate, the gyro's spin axis can only be horizontal with respect to the earth at one particular bank angle, i.e. at one particular True airspeed.
Whenever the gyro's spin axis is banked relative to the instrument case and the aircraft-- i.e. whenever the aircraft is turning-- the gyro responds to pitch rotation as well as yaw rotation.  For simplicity, consider a case where the gyro's spin axis is banked almost ninety degrees relative to the aircraft, which would never actually happen.  Now yaw rotation will barely be sensed at all, but any pitch rotation by the aircraft would be converted by gyroscopic precession to a roll torque on the gyro, tending to roll the gyro's spin axis to a steeper bank angle.
This effect is called "looping error".  One consequence of "looping error" is that any time the flight path is curving upward-- i.e. any time the aircraft is experiencing more than the normal G-loading for the bank angle-- the Turn Rate Indicator will read higher than the actual turn rate.
It appears that whenever the gyro's spin axis happens to be exactly horizontal with respect to the earth and the aircraft's flight path is not curving upward or downward, the deflection of the indicator needle will be exactly proportional to the turn rate.  (Citation needed.)  In essence, "looping error" is exactly compensating for the fact that the turn comprises of both pitch and yaw, rather than pure yaw.  However, as noted above, a turn rate indicator cannot be designed so that this happy situation occurs at all possible bank angles for any given turn rate.  Rather, in a steady-state turn at a given turn rate, this situation can only occur at one possible bank angle.  In many instances the gyro's spin axis will not be exactly level with respect to the earth.  So for any given actual turn rate, the indicated turn rate ends up being somewhat dependent on bank angle, which means that the indicated turn rate cannot be independent of True Airspeed.
It also appears that the stiffer the centering springs of the instrument, the less sensitive the instrument will be to transient "looping error" caused by pulling excess "G" in relation to the bank angle, but the less accurate the instrument can be over a wide range of True airspeeds in a steady-state turn, because the spin axis of the gyro will tend to stay more aligned with lateral axis of the aircraft (i.e. the wingspan) and less horizontal relative to the earth.
In a turn coordinator, with a symbolic moving airplane on the front of the instrument rather than a needle, the situation is even more complicated.  The orientation of the gimbal axis is different.  Rather than being oriented purely fore-and-aft, i.e. parallel to the aircraft's longitudinal axis, it is tilted (with the forward end raised and the aft end lowered) by about 30 degrees.  The purpose of this is to make the instrument sensitive to roll as well as yaw, to better signal the earliest stage of a developing turn or change in bank angle.
Apparently not all turn coordinator gyros spin the same direction.  If the gyro spins clockwise as seen from the right, with the top of the gyro moving away from the pilot, "looping error" is similar to what we see in the turn rate indicator, and if the gyro spins counter-clockwise as seen from the right, with the top of the gyro moving toward the pilot, "looping error" is reversed, so that the instrument under-reads the turn rate whenever the G-loading is excessively high for the bank angle.  Note that in the latter case, with the gyro spinning counter-clockwise as seen from the right, the symbolic airplane must be geared to tilt in the same direction that the gyro is tilted, not the opposite direction.  This means that in a steady-state coordinated turn the spin axis of the gyro actually is tilted away from level with the earth's surface, rather than toward level with the earth's surface.  In a steady-state turn, this would seem to make the displayed turn rate more dependent on bank angle, and therefore more dependent on True airspeed, than would be the case if the gyro spun the opposite direction.  For more on the direction of spin of the gyros in turn coordinators, and the resulting effect on the direction and magnitude of "looping error", see this related answer.
Any slight error in the turn rate indicated by a turn rate indicator or turn coordinator in a stabilized turn due to the calibration being only valid for one True airspeed is insignificant enough that it is rarely taken into account in actual practice.  This error is the clearly the least of a pilot's worries when flying in cloud.  But the answer to the question is that the calibration of any given turn rate index on any given turn rate indicator or turn coordinator can only be exactly accurate at one specific True airspeed.
Some related links--
How Gyroscopic Turn & Slip Indicator Works (You Tube video) -- in the cutaway, note how far the gyro's spin axis tilts to the side (rolls) in response to yaw rotation-- the tilt at full deflection appears to be about 40 degrees.  We can see that the tilt is in the direction that would bring the gyro's spin axis closer to vertical with respect to the earth during a normal coordinated turn.  By the fact that the gyro's spin axis rolls to the left when the instrument is yawed to the right, we can see that the gyro must be spinning clockwise as viewed from the right, so that the top of the gyro is spinning away from the pilot.
"How a Turn Coordinator Works-- Inner Workings"  -- Again we can see that gyro's spin axis rolls in response to yaw, and in this particular turn coordinator, the roll is in the direction that would bring the gyro's axis of rotation closer to vertical with respect to the earth during a normal coordinated turn.  Again, by the fact that the gyro's spin axis rolls to the left when the instrument is yawed to the right, we can see that the gyro on this particular turn coordinator must be spinning clockwise as viewed from the right, so that the top of the gyro is spinning away from the pilot.
Bristol Groundschool ATPL Training Material sample v6.1.5, published November 13, 2013 -- see page 3.13 regarding the dependency of the calibration of the turn rate indicator upon True airspeed.  See figure 3.16 on page 3.12 for an illustration of a turn rate indicator with multiple rate markings-- the typical British practice?
A link to an aviation website giving a rule of thumb for bank angle for a standard rate turn-- the rule of thumb is that the bank angle in degrees should be roughly .15 * (True airspeed in knots).  Reaffirming the idea that if the calibration of a Turn Rate Indicator or Turn Coordinator depends on bank angle, then it cannot be independent of True Airspeed.
https://www.pprune.org/tech-log/82721-turn-gyro-axis.html -- from an airline pilot's discussion forum-- not all the ideas here appear to be correct, especially the comment about the rotation about the gimbal axis being limited to 6 degrees of motion.
https://www.pprune.org/tech-log/528255-turn-rate-indicator-turn-coordinator-looping-error.html#post8164645 -- from the same forum as above--  a comment on the problems that can sometimes be posed by a Turn Rate Indicator's tendency to over-read under excess G-loading, and the greater problems that can be posed by a Turn Coordinator's tendency to under-read under excess G-loading.
Related ASE answers --
May turn rate indicators (and turn coordinators) be calibrated to be accurate at all airspeeds, or only for one? -- a good answer to what is essentially a duplicate question
During "partial panel" flying, in what ways is a turn rate indicator more useful than a turn coordinator? -- see all answers
In some "turn coordinators", does the gyro spin opposite to the direction it spins in a "turn rate indicator"? -- more on how the tilted gimbal axis of a turn coordinator allows roll sensing as well as yaw sensing, and also affects "looping error".  More on why if the gyro spins counterclockwise rather than clockwise as seen from the right, the instrument will under-indicate rather than over-indicate the turn rate whenever the G-load is unusually high for the bank angle.  Includes several links to videos and diagrams showing the internal workings of turn rate indicators and turn coordinators.
What is the formula for the bank angle required for a turn in line-abreast formation? -- calculating the required bank angle as a function of turn rate and True airspeed

First, way too long.  It was a pretty simple question, and the OP wouldn’t have asked if there wasn’t a basic understanding already that TAS affects the turn rate for a given bank angle.  A more concise (and correct) answer is called for.  Next, “Only at one particular True airspeed will the aircraft's actual turn rate exactly correspond to the reading of the turn rate indicator.”  Can you justify this statement?  Because if true, that ought to be published in every POH, right?  I will stop there for now.

Let us continue this discussion in chat.

@MichaelHall -- Deleting most of my comments.  Re "And why did you feel compelled to contradict a 4.5 year old correct answer with 14 upvotes?!! (15 now since I just added mine...)" -- because it was wrong, as explained in the present answer.  That's really all that need be said.  But see also https://aviation.stackexchange.com/a/65651/34686 .  (Not my own.)  As for the length, well, when one is refuting something that (almost) everyone "knows" to be true, sometimes one has to go into some detail.

Found a good "rule of thumb" for GA: airspeed divided by 10 + 5 for bank angle.  This was applied to fairly good agreement to a 172s standard turn rate bank angle of around 17 degrees (still depending on airspeed).  One would probably want to match cruising speed to turn rate (for that particular instrument) as much as possible.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


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What is the relation between aircraft stability and control?
:)
My teacher gave the task to figure out the common relation between stability and control. I know definitions and only thing I could find on the internet was "Control is directly opposed to stability" but I am not sure if I get it. Thank you for your answers :)

Simply put...
If something is very stable, it is difficult to get it to change attitude. If something is unstable it is easy to change attitude (not necessarily in a controllable manner).
Civilian aircraft tend to be very stable (easier, safer operation and more comfortable). Fighter jets tend to be (very) unstable for manoeuvrability. Some fighters are so unstable (F-16) that they need artificial stability (electronics) or no human would be able to fly it.

"Control is directly opposed to stability" -- yet stability is not always directly opposed to control.  Example-- if a hang glider "tumbles" due to getting too slow at the top of a loop, it becomes very difficult to control.  Consider also an aircraft that will enter an unrecoverable flat spin at the drop of a hat-- very difficult to control.

You are right.
Stability is the tendency of a system to return to its old state after a disturbance, be it a gust or a control input. It needs more control effort to change the state of a more stable airplane, such that stability can be measured by measuring control forces instead.
A famous example which is often mentioned in literature is the BE.2, a British two-seater which was designed as a training aircraft in 1912. As a trainer, it had ample stability, so its control responses were sluggish. That proved to be its undoing when war broke out in 1914. The BE.2 was easily outmaneuvered by dedicated fighters, so it had to be withdrawn from frontline service.
Quote from Wikipedia:

While it proved fundamentally unsuited to air-to-air combat it had a
  relatively low accident rate, and its high degree of inherent
  stability actually proved helpful in its artillery observation and
  aerial photography duties (although it also made it more difficult to
  maneuver quickly)

BE.2 (picture source)

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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Does "pendulum effect" apply to hang gliders or any aircraft?
In discussion of aircraft roll stability in FAA paragraph on dihedral effect and in keel effect and in Why are high wing aircraft more stable the question of pendulum effect arises.
This is usually described as center of gravity displaced laterally from below center of lift during a roll in a high wing airplane causing a rolling moment which restores wings to a level orientation.

source
In any explanation you give, please consider the case of a hang glider which depends on weight shifting to control roll and pitch. (Don't need to discuss pitch)

source

All of the links in your question seem to point to answers that say "no, it's a myth".  Are there any counterexamples?  Right now I don't see any controversy.

@BowlOfRed I think the hang glider is a counterexample.

What is usually understood as a "pendulum effect" is the normal (centered) fuselage contributing to the stability, i.e. creating a rolling moment by the virtue of being lower. This is a fallacy. But once you have misaligned lift and weight (hang glider), i.e. the lift vector not going through C.G., you'll have a rolling moment.

There is an aerodynamic effect (often called the Keel effect) that is sometimes called the "pendulum effect".  There is also the idea that a plane that "hangs" below a high wing is stable due to gravity pulling it upright.  I suspect that some things you find may be referring to one and some to the other.  Are you asking about one of these concepts specifically?  Wikipedia links it to the keel effect.

@BowlOfRed What I have read equates keel effect and pendulum effect, so both are in question.

The fallacy is to think that the hangglider pilot shifts the center of gravity: He can't, because of the conservation of momentum. Instead, he shifts the wing and with it the center of lift to the opposite side. All rolling motion is around the center of gravity so the weight has no pull on the rolling moment (pun intended).

@user207141 and PeterKampf and others-- all this attention to what is the immediate (before any roll) effect on the combined CG of glider plus pilot, when pilot exerts force with muscles to intentionally "shift" to side, as seen from a reference frame originally moving simultaneous with glider, is "interesting", but kind of peripheral to "pendulum stability".   In fact I'd even suggest that any discussion of pilot intentionally shifting to side is peripheral.

(ctd) Because it's obvious enough why the pilot's body intentionally shifting far to one side creates a roll torque.

@PeterKämpf and user207141 and others -- (ctd) A more interesting question is in turning flight, does the hang glider pilot have to exert any muscle force just to keep himself centered in the control frame, rather than falling to the "low side", and if so, is this evidence of some sort of "pendulum stability" effect? And would this same effect exist to any degree if the pilot relaxed all muscle force and allowed himself to swing freely? In the latter case, would it make any difference as to whether the flexible hang strap attached to the glider above or below the glider's CG?

For a proper discussion, we should first define what a pendulum is. Only then can be established if such an effect can exist in airplanes.
Let's base the definition on Wikipedia. It says that

A pendulum is a weight suspended from a pivot so that it can swing freely.

Maybe it is also worth to look closer what a pivot is: A thing on which something turns.
So the pendulum is fixed to a fulcrum which keeps it suspended and allows it to swing freely. The ideal pendulum has all its mass in its massive bob and, therefore, pivot and center of gravity are not in the same place. If the center of gravity and the pivot would fall together, a pendulum could only rotate but not swing. And that swinging motion is what the pendulum is all about.
Now for airplanes: Here we have no pivot. All rotation can only happen around the center of gravity. This is equivalent to the pendulum with no length which is no pendulum any more. But what about hang gliders? Using Pilothead's sketch let's consider how a hangglider starts a rolling motion and compare that to how a glider does it. The top two sketches show each in steady straight flight while the bottom two sketches show both initiating a roll to the right:

In both cases, a lateral imbalance is needed to create a rolling moment (red round arrow). While the pilot of the hang glider shifts the whole wing sideways, the glider pilot commands a lift difference between both wings using ailerons. Note the shift of the hangglider to the right and with it the center of lift in the lower left sketch: The center of gravity stays where it is due to the conservation of momentum while the pilot will shift slightly to the left.  In case of the glider, no such lateral shift of the craft occurs; instead the distribution of lift is shifted in order to create the imbalance. The effect is the same: A lateral displacement between weight and lift which causes a rolling moment. Rolling happens around the center of gravity (because of conservation of momentum again) and in all cases there is no displacement between the pivot point and the working point of the weight, because both are the same: The center of gravity.
A hangglider is only superficially different from an airplane because the center of lift is actively shifted by displacing the wing laterally instead of a change in lift distribution (neglecting the influence of the pilot mass on the rigging of each wing), but in all cases the vehicle will rotate around its center of gravity. Weight has no lever arm to that center of gravity, so there can be no pendulum effect. Or keel effect, fo that matter.

I've always understood that the keel effect is the roll-sideslip coupling due to pressure on surfaces above or below the cog (I.e. purely aerodynamic and not like a weighted keel). Is this not known as the keel effect, or is this not a real effect?

@Sanchises: Yes, that makes sense, but there is already the dihedral effect to occupy that slot. Granted, this name sounds as if only the wing is involved, but a fin or vertical tail (with a dihedral of 90°) shows the same effect, and for the same reason. I see no need to add yet another name, but admit that a real sailboat keel works in much the same way. In the Wikipedia article however, it is used as another name for the pendulum effect (but then it also confusingly talks of an aerodynamic side force).

Ah I suppose indeed a vertical tail "keel" is indeed just 90° dihedral. The wiki article is useless I agree. Would you consider the fuselage or pontoons dihedral (or anhedral for pontoons)?

@Sanchises No, both the fuselage and pontoons would have too little aspect ratio and too little lever arm to create much of a rolling moment. On the other hand, a low fuselage is letting the wing react like it had a bit more dihedral - that is about it for fuselages, however.

As an aside, modern hang gliders actually have a strong "aileron" effect that is an important part of roll control. The pilot's shifted weight actually pulls the "keel tube" to one side and the fabric of the left and right wings shift in an aileron-like manner.  But, that's peripheral to the main thrust of the question asked here.

@quietflyer: Absolutely correct on all counts.

@PeterKämpf "All rotation can only happen around the center of gravity." Airplane dont rotate about c.g. https://aviation.stackexchange.com/questions/83456/does-aircraft-allways-rotate-about-center-of-gravity-like-weather-vane-about-piv

@user207141 Accepted answers ≠ correct answers. Try to lean about conservation of momentum and don't leave thinking to others.

@PeterKämpf Dont agree with your statement that pilot dont change position of c.g.  at hang glider system, wing has mass as well. I explain this in my answer.

@PeterKämpf -- re "Now for airplanes: Here we have no pivot. All rotation can only happen around the center of gravity."  -- see my recent comments to TomMcW under my own answer-- re the fact that we can choose any arbitrary pivot point when analyzing torques.  That being the case, what does it mean to say that rotation can "only" happen about the CG?  I don't think there's any basis for such an assertion.  (Maybe grounds for a new ASE question?)  As a simple counter-example, consider a barrel roll, or a loop, or an ordinary turn.  I think that the assertion would only be true in a vacuum.

The pendulum effect is there in high wing a/c in theory but the effect is negligible because the moment arm of the center of mass is so short relative to lateral aerodynamic center. As with dihedral, for the effect to work you have to have sideslip, and the center of mass has to be below the lateral aerodynamic center of the sideslipping fuselage so that a rolling moment is created by gravity vs the lateral force.
Best way to imagine it is the take it to the extreme; imagine a rigidly installed weight on a solid pole extending 50 feet below the fuselage.  If the airplane slideslips, there will be some lateral lift force generated opposite to the sideslip direction centered somewhere on the fuselage, and with the center of mass way below that due to the weight on the pole, the center of mass is going to want to get below the fuselage.  On a normal high wing airplane, this effect is negligible if there at all.
Not for parasails however.  Pendulum effect provides all of a parasail's lateral stability.  If a parasail starts to sideslip, your ass wants to get below it because the center of mass is down where you are and the lateral aerodynamic center is up at the wing.  The pendulum effect on a parasail is so strong that they are able to bank by skidding even though the skid is induced by increasing lift on the inside half of the wing (it's the accompanying drag that's doing the actual turning).  In other words pendulum effect overpowers the opposite rolling moment of the lowered trailing edge.
On a high wing aircraft the main effect of the configuration is aerodynamic dihedral, which is the differential lift created by sideslip where spanwise flow is obstructed below the wing but not above by the T configuration, which tends to increase the lift of the low wing.  A high wing aircraft can have sufficient self-leveling effect from aerodynamic dihedral that it can get away with no geometric dihedral, and the wing is straight, although most include some geometric dihedral as well.
A good example of this is the CL-215 waterbomber.  The wings are straight and there is sufficient dihedral effect from the T wing placement to serve the roll stability needs of the aircraft's mission.   But, when the conversion to the turboprop CL-415 was done, the flat nacelles of the PW123 engines were discovered to have a blocking effect on spanwise flow above the wing in sideslip that was equivalent to extending the fuselage above the wing, killing a lot of the aerodynamic dihedral effect.  Any pendulum effect that was there wasn't significant (if anything, the center of mass of the 415 was lower because of the lighter engines vs the R2800s of the 215).
The fix for the CL415 was one of the biggest aerodynamic band-aids I've ever seen, those odd little wingtip extensions (they are NOT winglets) which create just enough dihedral effect in sideslip to restore what was lost from the addition of the fin-like nacelles.
https://upload.wikimedia.org/wikipedia/commons/9/9a/Canadair_CL-415_Filling.jpg
You see the opposite with high wing aircraft with sweep because there is a strong dihedral effect from the sweep itself. The combination of T configuration and sweep creates way too much dihedral effect.  So, just about all high wing airplanes with swept wings have anhedral to cancel out the excess.

All that you describe except for the last paragraph is dihedral effect. This is a completely aerodynamic effect and has nothing to do with gravity or a pendulum. What the question is getting at is whether gravity can cause a roll moment due to CoG being lower than the center of lift.

The first sentence answered the question.  There is an effect but it is too mild to be significant.  The rest was an explanation of why, and what "aerodynamic dihedral" is about, which is the real benefit of the high wing configuration from a lateral stability standpoint.  Seems pretty straightforward to me.

But some people dispute that there even is such an effect, no matter how small. I understand all the effects due to sideslip,  but does the fact that the cg is not directly below the center of lift create a roll moment? PK made the point that the lift vector is always radial to cg, therefore can create no roll moment, no matter how far away it is

Ok I see your point. I've edited my post.  See if that helps.

"The pendulum effect on a parasail is so strong that they are able to bank by skidding even though the skid is induced by increasing lift on the inside half of the wing (it's the accompanying drag that's doing the actual turning). In other words pendulum effect overpowers the opposite rolling moment of the lowered trailing edge."-- so they are deflecting their "ailerons" the "wrong" way!  Actually I've heard that before.  PS didn't human-powered Gossamer Albatross do same?

(In that case the issue being not pendulum effect but adverse yaw and resulting roll torque due mainly to differential airspeed between left and right wings.)

Pendulum effect don`t exist in aircraft, because lift force act perpendicular to wingspan, not vertical(like at flying balloon).
When pilot move to the left, c.g. of hang glider system shift  to the left in relation to the lift line, causing lever arm between gravity and lift force, that induced roll.
It is the same like you move very heavy cargo from center to the left  in transport aircraft during level flight, aircraft will roll to the left.
Position of c.g. of hang glider system  in reference frame of earth is irrelevant here, because we are interested only at c.g. of system in relation to lift line.
Position in space of c.g. of hang glider system cant not be at same position as it was when pilot stay at center, because lateral air resistance of pilot and wing are not the same.
If we neglect lateral air resistance, then c.g. of hang glider system stay at same place.(Imagine man in space hold rock of 20kg in outstretched hand, if man pull rock toward his body, center of mass system(men+rock) must stay at same position. Center of mass system must stay at same place,because you cant translate center of mass system with internal forces.)

While the pilot can shift his personal cg relative to that of the hang glider, he cannot shoift the cg of the whole aircraft sideways without violating the conservation of momentum. When the pilot shift his cg to the left, the cg of the hang glider is shifted to the right and the common cg stays where it was.

@PeterKämpf Do you agree if you shift cargo from center to the left in aircraft during level flight, aircraft will roll to the left?

@PeterKämpf "the cg of the hang glider is shifted to the right and the common cg stays where it was." Position of c.g. of hang glider system cant not be at same position as it was when pilot stay at center, because lateral air resistance of pilot and wing are not the same. If we neglect resistance forces, then c.g. of system stay at same place.

@user207141 If you shift cargo to the left, you need to push against something (Newton's 3rd law). This something is the airplane which will be moved to the right slightly. Now the wing is shifted right while the cg is shifted left by the same amount relative to the airplane. Thus, it will create more lift on the right side and roll left. Around the cg.

@PeterKämpf I agree with you,but only if you neglect  lateral air resistance which appears when pilot move to the left and wing move to the right. Imagine diver in water hold in hands flat plate of 2m x 2m, if he pull plate toward him, c.g. of system(diver+plate) will not stay at same position because flat plate has higher drag then diver.I know this effect in air is small,but still exist. Do you agree with my logic?

Retracting all comments, will incorporate relevant ones into revised version of own answer, or replacement-- but still say all the emphasis on whether/how combined CG of glider-pilot system moves when pilot weight-shifts is "barking up the wrong tree".  A more useful question is "would the pilot have to exert a muscle force just to keep his body exactly centered in control frame during banked turning flight, and if so, in what sense does the force or roll torque exerted by his hands on the control frame represent a contribution to roll stability that might be described as a 'pendulum effect'"?

(The "barking up the wrong tree" question could actually serve as a new ASE question!)

@user207141 This lateral speed is so small that you can completely ignore it. In effect, it results in a tiny sideslip angle for a short moment. Water is 800 times more dense and a 2x2m plate is much larger than the side area of a hang glider or the human below it.

@PeterKämpf I agree with this.. Why did you say this accepted answer is not correct, can post your explanation for this topic ? https://aviation.stackexchange.com/questions/83456/does-aircraft-allways-rotate-about-center-of-gravity-like-weather-vane-about-piv

@user207141 Conservation of momentum means that a force is needed to change the trajectory of a mass. Without an outside force, any rotation of a hangglider must be around its cg. To say that rotations around the cg are rare is simply wrong.

@PeterKämpf When side gust hit vertical stabilizer or when aircraft roll with activated ailerons why would rotate around c.g. if these are both outside/external forces? If we exclude couple force, isnt object only rotate around center of mass in absence of external forces?

@user207141 Strictly yes, but the gust will create a side force (resulting in a translational movement) and a moment (weathervane stability) which results in a rotation around the cg. Since the forces ahead and aft of the cg are nearly balanced, the side force will act very near the cg, and only the long lever arm of the aerodynamic surface with the highest lift gradient (the vertical tail) will make sure that the moment is turning the aircraft into the gust.

The "pendulum effect" as used in aerodynamic theory does NOT necessarily involve a weight that is free to swing. Maybe it's a poor phrase to describe what we are talking about, but it's the one that has become common. It's also a bit misleading because it implies that gravity itself is somehow exerting a direct roll torque on the aircraft.  That is not actually the case-- gravity or weight acts at the CG and so does not exert a direct roll torque.  Still, a low CG placement DOES tend to lead to a stabilizing roll torque that is much like the roll torque contributed by dihedral, sweep, etc.  All of these effects contribute a stabilizing "downwind" roll torque in the presence of sideslip.  All of these effects could be said to shift the aircraft's overall "effective dihedral"-- the relationship between sideslip and "downwind" roll torque-- in the more positive or less negative direction
Consider free-flight model airplanes with the wing on a pylon high above the fuselage.  Consider paragliders which have a strong anhedral geometry to the arched wing and yet are generally quite stable roll in roll -- as illustrated by countless stories of both deliberate and accidental cloud-flying in such aircraft with minimal instrumentation and yet with acceptable outcomes.
Note that the multiple lines connecting the paraglider pilot to the wing essentially act like fixed struts due to the triangular geometry involved.
The key to understanding the "pendulum effect" lies in understanding that a turn generally involves some sideslip (for reasons that are not simple), and that during a sideslip, the aircraft's drag force has a sideways component relative to the aircraft's longitudinal axis, and we also generate aerodynamic sideforce ("sideways lift", acting perpendicular to the lift and drag vectors) as the airflow strikes the side of the fuselage, vertical tail, etc.  Any sideways force that acts above or below the CG will contribute a roll torque.
In a paraglider, the same anhedral wing geometry that must contribute some amount of destabilizing "upwind" roll torque during a sideslip due to the difference in angle-of-attack between the left and right halves of the canopy or wing, also exposes a great deal of surface area to the sideways flow, high above the CG, contributing a stabilizing "downwind" roll torque-- the "pendulum effect".  Obviously in a paraglider, the latter dominates over the former.
High-winged aircraft benefit from enhanced roll stability due to the "pendulum effect", though there is also an additional "downwind" roll torque created by interference between the fuselage and the wings.  The latter may be absent if the wing is mounted on struts high over the fuselage-- the "parasol" configuration.
In a hang glider, the pilot hangs from a flexible strap typically connected near the aircraft CG.  In such a case, a "pendulum effect" is only present when the pilot locks himself into place with his arms-- i.e. when he uses his muscles to make a roll input.  When he is hands-free, his weight acts at the CG and there is no "pendulum effect", even though during a sideslip, his body does tend to displace slightly (a few inches) toward the "upwind" side of the control frame, just as a slip-skid ball would.  Note that the pilot's tendency to swing slightly to the upwind side of the control frame during a slip, is simply a reflection of the sideways component of the wing's drag force plus the aerodynamic sideforce generated by the wing -- if these were zero the pilot would have no tendency to deflect toward the upwind side of the control-frame, and a slip-skid ball would remain centered. (In fact, in such a case the wind would actually blow the pilot toward the other side of the control frame-- the downwind side-- during a slip.  The pilot's body would act more like a yaw string than a slip-skid ball!  This is not what we observe in practice.)
In this answer, except where specifically stated otherwise, we'll consider the hang glider in the "hands-free" case-- i.e. when the pilot is exerting zero muscle force.  The same dynamics do also affect the control inputs (muscle force) the pilot must exert to get a given result (e.g. roll rate), but we won't explore that very deeply in this answer.
On some older designs, the pilot's "hang strap" connected to the hang glider several feet below the "keel tube"-- in this case the pilot's weight acted well below the CG of the glider, and so the sideways aerodynamic forces generated by the wing during a slip did in fact contribute a stabilizing "pendulum effect".  In such a case it's equally valid to look at the pilot and the glider as separate bodies, and note the roll torque generated by the sideways pull of the pilot's hang strap on the glider, or to view the glider and pilot as one system (with the mass of the pilot considered to be located at the point where the "hang strap" connects to the glider), and note the roll torque generated by the sideways aerodynamic forces acting above the CG of the whole system.
On many modern hang gliders, the pilot's "hang strap" actually connects to the glider partway up the kingpost, or on kingpost-less gliders, on a little stub sticking several inches above the keel tube.  In this case the pilot's weight acts above the CG of the glider, so the interaction between the sideways aerodynamic forces in a slip, and the mass of the pilot, contributes a destabilizing "upwind" roll torque-- an "anti-pendulum" effect.  This is done to increase maneuverability.  Hang gliders experience substantial sideslip due to adverse yaw while rolling, so excessive "effective dihedral"-- contributing excessive "downwind" roll torque in the presence of sideslip-- is very undesirable and greatly limits the roll rate that can be achieved.
Note that the gull-shaped wing we see in many hang gliders-- especially in a view of the trailing edge-- contributes a dihedral geometry to the inboard part of the wing and an anhedral geometry to the outboard part of the wing.  Even if the net result in terms of a pure dihedral effect is zero-- which may or not be the case-- this type of design does increase the total amount of surface area exposed to the sideways airflow during a sideslip.  So the sideways aerodynamic force component during a slip will be larger with such a design, than with a completely flat wing.  This is probably best described as an unintended consequence of a wing shape that evolved for other reasons.  In some older hang gliders that had vastly more sail "billow"-- more "arch" to the trailing edge-- than today's designs, the pilot probably experienced a much larger tendency to "fall" toward the low or "upwind" side of the control frame during a sideslip, due to the increased side area exposed to the airflow.
As noted above, the relationship between bank angle, turning, and slipping is complex.  It is driven in part by the increased drag experienced by the outboard wingtip in turning flight, due to its higher airspeed.  It's a misconception to think that banking automatically generates a slip simply because weight now has a sideways component in the aircraft's frame of reference.  It's also a misconception to think that banking automatically generates a slip simply because the lift vector is now tilted relative to the earth and the wing's lift vector now has a horizontal component-- that is true in any banked turn, slipping or not.  Sometimes yaw rotational inertia may play a significant, albeit transient, role in causing sideslip.  In actual flight, in many aircraft (including hang gliders) we can observe that slip is driven overwhelmingly by roll rate, and to a much smaller extent by yaw rate.  One example of a case where we can often see substantial sideslip with no roll rate, is as we go over the top of a wingover, with a 90-degree bank angle, with no rudder input.  A complete exploration of exactly what maneuvers will involve sideslip, and to what extent, and why, is far beyond the scope of this answer.

’exposes a great deal of surface area to the sideways flow, high above the CG, contributing a stabilizing "downwind" roll torque’ Can you explain this any better? I'm not understanding what you mean here.

Imagine an aircraft with totally flat wings, as viewed head-on.  Now imagine an aircraft with gull-shaped wings, as viewed head on.  Imagine that the wingtips are no higher or lower than the wing roots, in both cases.  Won't the gull-shaped wing have more area exposed to the sideways flow during a sideslip, and thus generate more aerodynamic sideforce, than the flat wing?  I guess by "sideways flow" I'm thinking specifically of the component that goes from one wingtip toward the other-- not necessarily parallel to the horizon, if the a/c is banked.

It would produce more drag. Can't figure out where the side force would come from. Not saying it's not there, just saying I don't see what would provide it.

Isn't intuituve that a sideways wind component blowing against that big arched shape would tend to "blow" it in the downwind direction in a manner that would not simply involve a redirection, with no amplication, of the usual drag vector?  OK I see your point, do we really have a sideforce, or do we just have an amplification, as well as a redirection, of the drag vector?  Have to think about that.

Any force with a moment arm about the cog will have a rolling effect. That is entirely aerodynamic and has nothing to do with the direction of gravity, which is what a pendulum is affected by.

Let us continue this discussion in chat.

Well, now it sounds like you are commenting on whether or not "pendulum effect" is really an appropriate term for what my answer focussed on.  I never at any point said that paraglider would automatically tend to respond to the gravity vector even in the absence of sideslip.

This answer mentions "gull wing" configuration, with dihedral inboard, and flat or anhedral outboard-- here's the most extreme example of that configuration that I've ever come across -- !! -- https://aviation.stackexchange.com/questions/21063/what-advantages-come-with-the-wing-on-the-boeing-bird-of-prey?noredirect=1&lq=1

@TomMcW -- (some years ago) you stated "Any force with a moment arm about the cog will have a rolling effect. That is entirely aerodynamic and has nothing to do with the direction of gravity, which is what a pendulum is affected by. "  Well, yes and no.  There's an interesting point here that goes rather deep.  To analyze the torque acting on a body, we are free to choose any arbitrary pivot point.  If we choose the CG for our pivot point, then the apparent weight vector (which in flight, is really the G-load vector, and if we are "pulling G's" or slipping/ skidding is different than the

(ctd) and if we are "pulling G's" or slipping/skidding is different than the actual weight vector), plays no role in our torque calculation.  But if we choose some other pivot point-- e.g., in some cases it might be convenient to pick as our pivot point the point where the wing's lift vector acts-- then the apparent weight vector, acting at the CG, does need to be included in our torque calculation.  Both approaches are valid.  Any conversation about the "pendulum effect" which overlooks this, will be doomed to go in endless circles--

(would have posted this to the chat room, but it is frozen for lack of activity.)--

Question is still apparently seen as "unresolved" after 4 years?  May need to post a whole new answer --

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How Jozza Bombsight worked?
I'm looking some information about the so called Jozza Bombsight used for example on board of the Savoia-Marchetti SM-79 during WWII. Thank you

Presumably this is a reference to the Italian WW2-era airplane "Savoia-Marchetti SM-79" (see for example https://www.historynet.com/damned-hunchbacks.htm).  You might want to add some of those details to clarify your question.

Thank you, I was interested   in how was for the pilot using it. I found many informations for example about the Bombsight on the B17 but nothing about the JOZZA...thank you,


text:
If in a coordinated turn, the horizontal lift vector is equal to the Centrifugal force. Then how is the aircraft still turning?

Image source
How does the Aircraft continue to turn when the both the Horizontal component of lift and the centrifugal force are equal?

https://thebackseatpilot.com/pages/aerodynamics#:~:text=AERODYNAMICS%20OF%20A%20TURN&text=The%20VCL%20is%20opposed%20by,maintains%20a%20level%2C%20coordinated%20turn.

Re the link above-- wow, the diagram labelled "Left bank, no back pressure" makes even worse errors.  Implies that the amount of back pressure you apply affects the vertical component of lift but not the horizontal component.  Fails to recognize that if true, this would create a skid (net aerodynamic force pointed to inside of turn, not "straight up", in a/c ref frame.)  Completely bogus.  In the hang gliding world an argument is sometimes made that inadequately pitching the nose up while turning creates a slip rather than a skid.  That's bogus too.  Slip/skid is about yaw not pitch.

Try it-- enter a turn and change the back pressure. You'll see little effect on the ball.

Just noticed that this question is a duplicate of https://aviation.stackexchange.com/questions/38040/how-can-an-aircraft-turn-if-the-horizontal-force-component-is-zero?rq=1 .  Some well-thought-out answers were posted there.

Related -- https://aviation.stackexchange.com/q/79338/34686

The forces in the first pictures would be as shown if you are observing situation from reference frame connected to the turning airplane. Then, in this reference frame, the airplane is not moving and therefore neither turning. Other forces, gravity and interaction with air molecules remains in place, so in order to make the description consistent, you need to add centrifugal force.
If you observe the situation from ground, there is no centrifugal force (because this fictitious/inertial force appears as a consequence of applying laws of motion in rotating reference system) and horizontal component of lift makes observed trajectory circular.
Edit2: The later two images are not complete. Forces has to be always in balance if viewed from airplane's reference frame. They are missing part of aerodynamic force generated by side-wise motion through air, which is definition of slip and skid (you can call it drag or airframe-lift). See @quietflyer's answer for more details on this.

Re "(if it should be a view from airplane's reference frame, CF and horizontal lift has to match always)" -- well, not if "horizontal lift" is only meant to include the horizontal component of the wing's lift and not the horizontal component of the aerodynamic sideforce from the airflow striking the side of the fuselage.  See my answer for more--

@quietflyer: True, I have realized this later, but had not found time to add it to my answer.

All fictitious, (non-inertial, psuedo forces), simply represent the acceleration of the frame of reference you are doing your analysis in. Doing your analyis in any accelerating, non-inertial frame invalidates Newton's kaws, UNLESS you add in an extra, fictitious force to represent the acceleration of the frame itself.

How does the Aircraft continue to turn when  both the Horizontal component of lift and the centrifugal force are equal?

Those pictures are not 100% correct.
To enter and keep a turn, a force toward the centre of the curve has to be created. In an airplane this is achieved by tilting the lift laterally, like in the following picture (source) where the airplane is turning left (as seen from the front):

The vertical component of the lift balances the weight out while the horizontal component keeps the airplane in a turn. This horizontal component is called centripetal force. The higher the centripetal force is, the steeper the turn is.
End of the story.
So what about the centrifugal force? Let's make an everyday comparison with what happen in car that accelerates. Due to the traction force the car gets accelerated forward. But what you experience as a driver/passenger is actually a backward force (aka inertia) pushing you against the seat. This is exactly the same as for our airplane: the centripetal force makes it bend leftward but what you experience is actually an opposite, rightward, centrifugal force. Those two forces have the same magnitude but opposite direction. Anyway is only the centripetal force that keeps the airplane in a turn. The centrifugal force is the inertia "felt" by the pilot/passenger and acting on all the non-lifting surfaces of the airplane.

The centrifugal force is produced by the act of turning.  If you weren't turning, it would be zero.  If it were zero, you would have an imbalance of forces which would cause an acceleration that would cause you to turn.
Properly, the centrifugal force is not really a force, it is an acceleration.
$a=V^2/r$
Recall that we can write $F=m\,a$ where forces balance out accelerations.  So, the horizontal component of lift needs to balance out (or it causes) the acceleration in the turn.
As @Martin said, the second two images are bogus and the source that generated them doesn't understand what is going on.

I do think "force is balancing out acceleration" formulation is misleading. Force is causing acceleration. In any reference frame. The problem with centrifugal force is that it is result of choosing accelerating frame of reference. So you could say, in some sense (this particular) force is "balancing out" acceleration of frame of reference (even if it is bit unusual wording), but frame of reference is no real object. For any physical object, force is causing acceleration, not "balancing" it.

@Martin Fair point -- I thought about wording things differently, but then I started to think about the acceleration due to gravity -- which we have in (what we consider) an inertial frame -- but that we often  freely replace with a force (weight).  I meant balance out to really imply algebraic equivalence.  And although the force undoubtedly causes the turn, when analyzing the problem, you may well observe the turn in order to then infer the force.

Acceleration causes Gravity, not the other way round,  or perhaps a better way of stating it is that they are the same thing. Think about it, how do we measure, or detect acceleration? With a G-meter. When we let Gravity accelerate us, (in free-fall), a g-meter will indicate zero.

I shouldn't have brought gravity into it.  It is a fundamental force of nature that is a body force (acts on all the mass simultaneously) rather than a point, pressure, or shear force that we're more used to thinking about.  Gravity is a force field.  If we are free to fall, then we accelerate at a fixed rate (due to gravity).  However, the acceleration does not cause it (the way an accelerating frame causes a force).  For if we can't fall, we still feel the force of gravity.  So a mystical force field causes the body force of gravity F=mg.

Centrifugal force is a fictitious force, i.e., it is not real, it is added by engineers only to make the F=ma equation work in what is called a non-inertial or accelerated frame of reference. They can also be referred to as Inertial or Psuedo Forces. It is the same concept as trying to do F=ma for motion in a room on a rotating space station that has artificial gravity due to the rotation. In that frame of reference, the cup of coffee on the table is not accelerating, it is motionless. To make the F-ma equation work, (to make the force F and the acceleration be zero we need to include a fictitious Centrifugal force to oppose and balance the centripetal force. Other forces which fall into this category include Coriolis force and Gravity.
That is exactly what is going on for the diagrams you are looking at for an aircraft in a steady state turn.
What is not addressed, (and rarely explained in basic flight training), is that Newton's rules of motion (F=ma for example), are only valid in an inertial or non-accelerated frame of reference. Unless you are in free-fall (experiencing zero acceleration), you are in a non-inertial or accelerated frame of reference, and some fictitious force must be added to your analysis to compensate for the acceleration of the frame of reference in which you are doing your analysis.
In addition, it must be remembered that everything is relative. Not just velocity. Most of us will agree that all velocities must be measured relative to some reference velocity. How fast is the earth moving? well, there is no correct answer unless you also specify "with respect to what"?  What is not generally understood is that the same concept also applies to acceleration.  Sitting in your chair in your living room, we are in fact in an accelerating frame of refence (as we feel acceleration), upwards. That is why Einstein's famous "thought experiment" is so illuminating. In it he explains that there is no way that someone can tell the difference between that situation, and being in a room in a spaceship in deep space that is accelerating with a rocket engine thrust at 32 feet per second squared. These two situations are not just similar, they are equivalent, (identical).  In both cases you are in a non-inertial, accelerating frame of reference, and a fictitious force must be added to any analysis to compensate for the intrinsic acceleration of the frame of reference to make any F=ma analysis equation balance properly.
Heree is a graphic from Misner, Wheeler & Thorne's Gravitation that illustrates this:

Note: centrifugal force is for most purposes as real as for example gravity. The fact that it appears because of choice of frame of reference does not make its effects less real. (You are actually telling just this in the last paragraph mentioning indistinguishably of gravity force and accelerating-frame forces.)

Actually, Gravity, as understood in the physics world, is also a "fictitious" force, as it is, according to General Relativity, (now pretty much accepted, verified and considered to be the current "correct" view of what gravity is),  is not a force at all, but just our perception of the straight-line path in space-time taken by objects that are not accelerated at all, because we are looking at them in an accelerated, non-inertial frame of reference, exactly as when we might look at the path of a dropped object in a rocket ship accelerating from thrust.

And saying that the effects are more or less real or not real, when the choice of which frame of reference you choose to look at them in totally changes how you see them is a bit foolish. It betrays a lack of understanding of what the issue is. Bottom line, If you can feel the force, it's real, if you can't then it is fictitious. Consider this, the only time we say we can "Feel" Gravity is when we are not accelerating. When we are accelerating due to gravity, (free fall), then we don't feel it. If this doesn't bother you a bit, then you don't understand this.

There are countless videos explaining this on the internet.  Here's a few... https://www.youtube.com/watch?v=XRr1kaXKBsU&pp=ygUfZ3Jhdml0eSBpcyBub3QgYSBmb3JjZSBlaW5zdGVpbg%3D%3D  https://www.youtube.com/watch?v=W3Jo6OCbuRk&pp=ygUfZ3Jhdml0eSBpcyBub3QgYSBmb3JjZSBlaW5zdGVpbg%3D%3D  https://www.youtube.com/watch?v=W3Jo6OCbuRk&pp=ygUfZ3Jhdml0eSBpcyBub3QgYSBmb3JjZSBlaW5zdGVpbg%3D%3D

The experience that created understanding for me was a diagram in a text by Wheeler, on page 33 of   https://ia800101.us.archive.org/24/items/GravitationMisnerThorneWheeler/Gravitation%20Misner%20Thorne%20Wheeler.pdf. (added to my answer above), In it he shows how the measured curvature of the geodesic (straight-line) path of a thrown baseball, a bullet fired from a gun, and a beam of light, all have exactly the same curvature. It's because the perceived is not a property of the moving object, it is a property of the acceleration of the frame of reference we are measuring things in.

Your quite natural, human bias towards considering that when standing or sitting on the surface of the earth, you are maintaining a constant velocity, and not accelerating, is due to the wrong assumption that there is a preferred absolute frame of reference, (that of the earth) and that acceleration is not relative, but absolute, and must always be measured relative to the Earth frame, and not to any other. In fact, GM says that the laws of Physics must be identical in all frames, and that Newtons Laws as he defined them are only valid in inertial or non-accelerated frames.

Charles, I know I won't win any argument with you on this point, but I'm with @Martin in terms of having an opinion that referring to anything we can directly experience as "fictitious" only serves to obfuscate common sense understanding for a majority of people.  (because why would you wear a G-suit to counter a "fictitious" force?)

Hah!  But we don't wear g-suits to counter the force of gravity! We wear them to counter the aerodynamic  force of Lift! And the fact that you confuse the two makes my point.. Even if you want to consider gravity a real force, it is never greater than 1 G, and you don't need a g-suit for that. Do you wear one walking around in your kitchen?

And another quick point here. Your intuitive comment about   "anything we can directly experience" also betrays the essence of this misconception.  You do NOT experience the force of gravity. Even before physicists understood that gravity is not a force, but just the curvature of space-time, when they thought that it was a force, they understood that this gravitational force was not, and could not, be felt by anyone, as it affected every atom in your body equally. (except for the edge case represented by extremely strong tidal forces - also fictitious).

But is it more correct to say that we wear G-suits to counter lift, or the effect of lift?

Yes, in a way, you are correct. We wear g-suits to prorect against the physiological effects of acceleration, (as measured in any zero-g inertial frame), which is indeed the resuIts of all real forces, (mostly lift), acting on the aircraft

@CharlesBretana: this is starting to be more philosophical (my note was actually only  about choice of words anyway), but real=feel-able is not good universal measure neither, because feeling depends on particular organs/biology. We are actually not sensing absolute force, but difference over our body parts. You would definitively feel gravity in (strongly) non-homogeneous field despite being (your c.g.) in free fall. And same goes for other "fictitious" forces. (Beletry link: Neutron Star by Larry Niven :-) )

@Martin, it's not philosophical at all. This misconception is the source of most of the confusion about this topic. And understanding this eliminates all inconsistencies, and makes everything make sense.  And you don't feel gravity in non- homogeneous fields, you feel tides, which are the differentials or divergence of the geodesics.

But Neutron Star is an awesome novel!

Philosophical in sense what words "real", "to feel" etc. mean. There is definitively more ways to look at situations and whatever fits your "worldview" and gives correct prediction of objective outcomes is as good way as another. :)

@Martin, if you believe everyone has a right to their own reality, and their own definitions, then of course.... but then why engage in discourse?  True - that each of us may understand any word to represent a different concept to some degree, but this is not a black and white, on/off thing, it's along a spectrum, and we need to use words in a minimally consistent way to communicate rationally.

Again, I knew ahead of time I had no hope swaying you, I just think referring to an effect we can experience directly as fictitious is about as useful as man’splaining to your wife that her remark about feeling “normal stress” from preparing to host your family for Thanksgiving this year is incorrect, because normal stress is when a force acts perpendicular to the surface of an object.  If feels very pedantic to me…  Cheers.

But again, you keep referring to "an effect we can experience directly". I have made it pretty clear, crystal, in fact, that what you are experiencing is not gravity but the aerodynamic force of lift on the wing.

Those diagrams, versions of which have been reproduced in many different flight training "ground school" materials, including some published by the FAA, are extremely misleading, and contain errors and omissions.
Slips and skids are not characterized by an "imbalance" between the pseudoforce called "centrifugal force", and the horizontal1 component of the net aerodynamic force generated by the aircraft.
Rather, slips and skids can be said to be characterized by an imbalance between the pseudoforce called "centrifugal force", and the horizontal component of the wing's lift force.  (This isn't the simplest or most intuitive definition of a slip or skid, but it is a valid one.)
Since the pseudoforce called "centrifugal force" is actually just a mirror image of the horizontal component of the net aerodynamic force generated by the aircraft, the above statement is really just an overly complicated way of saying that in a slip or skid, something other than the wing is generating an aerodynamic force which has a horizontal component.
That "something" is the fuselage.  In a slip or skid, the airflow is striking the side of the fuselage, which generates a real aerodynamic sideforce, oriented perpendicular to the wing's lift vector.   Because the fundamental defining characteristic of a slip or skid is that the nose of the aircraft is not aligned with the actual direction of travel through the airmass, in the yaw axis.
If the aerodynamic sideforce generated by the airflow striking the side of the fuselage were included in the diagrams attached to the question, the horizontal component of the net aerodynamic force generated by the aircraft would be exactly equal in magnitude (and opposite in direction) to the "centrifugal force" vector in every case.
If the vectors labelled "HCL" are supposed to represent the horizontal component of the net aerodynamic force generated by the aircraft, including the horizontal component of the aerodynamic sideforce generated by the airflow striking the side of the fuselage, then they are drawn incorrectly.  For a given bank angle, the horizontal component of the net aerodynamic force generated by the aircraft is larger in a skidding turn than in a coordinated turn, and is smaller in a slipping turn than in a coordinated turn.  If the vector labelled "total lift" is supposed to represent the total aerodynamic force generated by the aircraft, including the sideforce contribution from the airflow striking the side of the fuselage, then the coordinated turn is the only case where the "total lift" vector should be exactly "square" to the wingspan.  In all cases the "centrifugal force" vector should be a mirror image of the vector representing the horizontal component of the net aerodynamic force generated by the aircraft.2
On the other hand, if the vectors labelled "HCL", "VCL", and "Total Lift" are only supposed to represent the horizontal, vertical, and total components of the wing's lift vector, then the diagrams become very confusing because the vector representing the aerodynamic sideforce generated by the airflow striking the side of the fuselage has been entirely omitted.  In this case the vectors labelled "centrifugal force" should still appear as described in the above paragraph, but in the slipping and skidding cases, the vector labelled "HCL" will no longer be a mirror image of the vector labelled "centrifugal force".  For a given bank angle, the wing's total lift vector, and therefore the horizontal and vertical components of the wing's lift vector, are all slightly smaller in a slipping turn than in a coordinated turn, because the aerodynamic sideforce generated by the airflow striking the side of the fuselage supports a small portion of the aircraft's weight.3  Similarly, in a skidding turn, the wing's total lift vector, and therefore the horizontal and vertical components of the wing's lift vector, are all slightly larger than in a coordinated turn, because the aerodynamic sideforce generated by the airflow striking the side of the fuselage contains an earthward component that must be opposed by the wing's lift vector.  These differences should be apparent in the vectors labelled "VCL" and "Total Lift" as well as the vectors labelled "HCL", and the vector labelled "Total Lift" should be "square" to the wingspan in all three cases.
Obviously the diagrams would be greatly improved by changing to an airmass-based reference frame rather than an aircraft-based reference frame, so that the "centrifugal force" vector could be entirely discarded, and by also including by the aerodynamic sideforce vector generated by the air striking the side of the fuselage.  There's no need to break things into horizontal and vertical components-- just show the wing's lift vector and the aerodynamic sideforce vector from the airflow striking the side of the fuselage.  These two vectors are oriented perpendicular to each other.  In a coordinated turn, there is no airflow striking the side of the fuselage, so the aerodynamic sideforce vector is zero, so the net aerodynamic force acts "straight up" in the aircraft's reference frame.  In a slip or a skid, the aerodynamic sideforce vector is not zero, and so the net aerodynamic force does not act "straight up" in the aircraft's reference frame.  End of story.  (If desired, an "apparent load" vector could be added, which would always be the mirror image of the vector sum of the wing's lift vector and the aerodynamic sideforce vector from the airflow striking the side of the fuselage.  Only in the case of the coordinated turn, where aerodynamic sideforce is zero, would the "apparent load" vector be exactly "square" to the wingspan.  And now-- keeping in mind that the weight vector makes no contribution to the "apparent load" vector-- we understand why the inclinometer ball behaves as it does in slipping, skidding, and coordinated turns.  And we've come to this understanding without ever invoking some sort of hypothetical "imbalance" between "centrifugal force" and some other force.  It all boils down to the question of whether the airflow is, or is not, striking the side of the fuselage and generating an aerodynamic sideforce vector.)
This concept was also explored in this related, highly upvoted question-- What is missing from these diagrams of the forces in slips and skids?
But as to your specific question:

If in a coordinated turn, the horizontal lift vector is equal to the
Centrifugal force. Then how is the aircraft still turning?

Because when we include the pseudoforce called "centrifugal force" in our vector diagrams, we are using an aircraft-based reference frame rather than an airmass-based reference frame or ground-based reference frame.  Since the aircraft can't accelerate relative to itself, the net force in the aircraft-based reference frame will always be zero, whether the aircraft is turning or not.  In the aircraft-based reference frame, the fact that the centrifugal force vector exists at all is actually evidence that the aircraft is turning.  But for teaching purposes, using the aircraft-based reference frame (and therefore including the "centrifugal force" vector whenever the flight path is not linear) is arguably an inferior approach to using the airmass-based reference frame (and therefore not including any "centrifugal force" vector.)
Footnotes:

In this answer, when we talk about "horizontal", we mean as seen looking at the aircraft in a head-on view.  We're not referring to fore-and-aft forces such as thrust and drag.

Note that in the diagrams attached to the question, the illustrator chose to make the "total load" vectors in the "skidding turn" and "coordinated turn" identical, and to make the "HCL" (horizontal component of lift) vectors in the "slipping turn" and "coordinated turn" identical. That's all kind of random and makes no sense.

For an extreme case, consider sustained linear "knife-edge" flight at a 90-degree bank angle-- here the aerodynamic sideforce from the airflow striking the side of the fuselage is doing all the work of supporting the aircraft's weight, and the wing's lift vector is zero.

Re "In a coordinated turn, there is no airflow striking the side of the fuselage, so the aerodynamic sideforce vector is zero, so the net aerodynamic force acts "straight up" in the aircraft's reference frame."  -- maybe "in the aircraft's coordinate system." would be a better way to end this sentence.  The reference frame we are using here is that of the airmass, not the aircraft.  (Hence no need to include "centrifugal force".)

Regardless of whether we are taking the approach outlined in the 7th paragraph ("If the vectors labelled "HCL" are supposed to represent the horizontal component of the net aerodynamic force generated by the aircraft..."), or the 8th paragraph ("On the other hand, if the vectors labelled "HCL", "VCL", and "Total Lift" are only supposed to represent the horizontal, vertical, and total components of the wing's lift vector..."), it makes no sense whatsoever for the HCL vector in the "skidding" case to be smaller than the HCL vector in the "coordinated" case.  It should be larger.

Likewise, at least if following approach outlined in 7th para, in the skidding case the Total Lift vector should be tilted further to the inside of the turn than in the coordinated case, not closer to vertical.  And the Total Load vector should reflect this, being a mirror-image of the Total Lift vector.  This set of diagrams really is completely flawed.

I'm gonna go out on a limb and suggest that all of these answers seem to be answering a different question than "why does the airplane turn" when the bank is such that the acceleration in the airplane's rest frame points straight down (a coordinated turn). A ball on a string swung over your head moves in a circle without any aerodynamic forces. So there's something else going on.
Part of the problem is conflating turn = "move in a circle" with turn = yaw.
You can have a banked airplane where the wing lift vector points off to one side but the airplane doesn't turn. Every PPL candidate has to demonstrate a slip to landing, which is exactly that - the rudder input exactly opposes the yaw tendency created by the horizontal relative wind component.
What that highlights is that the yaw is caused by the side-force of the relative wind on the tail. The horizontal relative wind at the tail is caused by the horizontal lift component created by the bank. In a constant rate turn the rate of yaw must match the rate of turn, but that doesn't mean the airplane's nose has to be pointed along the tangent of the turning circle.
If the apparent "g" in the airplane frame is not straight down, that will cause a horizontal wind component on the tail which will yaw the plane. If the horizontal "g" points to the inside of the turn, the effective wind will yaw the nose to the inside of the turn & vice versa. That's why you can make a turn without using any rudder - the "g" to the inside will create enough yaw to keep the plane turning. But it's less aerodynamic than if you get all the yaw from having inside rudder input instead of depending on inside "g". So "coordinated turn" = "apparent g straight down".

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Newton’s first law.  If the forces are perfectly in balance then the aircraft will continue to turn at a constant rate.

No, that's not true. With forces in balance (as seen from the given reference frame), the movement will continue along straight line with constant speed.

An object moving at a constant velocity is not accelerating, all forces considered.  Notice, in a turn, fuel input (power) must be increased, due to increased drag.  But once in the turn, fuel burn is constant.

@Martin ok the principle is more succinctly described by conservation of angular momentum but if you consider the aircraft as a collection of individual components then 3rd law applies; the frame of reference is the aircraft as a whole, it ‘sees’ the universe rotating around it and there’s no resultant force that would change it.

Anytime you attempt to analyze a scenario within, or with reference to, a non- inertial or  accelerated Frame of reference, (one in which a g-meter will show any non-zero value), you must include some non-inertial or fictitious forces that represent the acceleration of the frame of reference itself or Newton's laws will not work - they will no longer be valid.


text:
Does AOA become 0 after the aircraft flies in a particular attitude for sometime as the relative airflow has changed and how does it affect Stall AOA?
As AOA is the angle between relative airflow and Chord Line. Then after you increase the angle of attack and fly at that angle of attack for some amount of time and the relative airflow changes accordingly. Does the AOA then become ZERO. If it does does it virtually mean that the we have more cushion to reach the critical angle of attack and stall the aircraft?
I have attached an image that shows the relative airflow changes as the attitude changes. It is from the Transport Canada Flight Training Manual.

No, the AOA does not become 0. The relative airflow does not change over time.  Where did you get this idea from?  Did you read this somewhere?

I have attached an image from Transport Canada FTM. Please have a look. Maybe this picture is misleading and I have misunderstood the explanation.

Those Transport Canada images are misleading because they have led you to believe something that is not true. Aircraft RARELY climb or descend with 0 AOA, and even in cruise flight there is usually some AOA.

As you have correctly written, the "AOA is the angle between relative airflow and Chord Line". In all the three cases of your picture the AoA seems to be just the same. What's you doubt exactly?

The image shows that the direction of the relative airflow changing with the attitude of the aircraft or atleast it looks like that. That led me to believe that Relative Airflow changes as you pitch up or down.

@MikeSowsun and Nish -- I would agree with the statement that the direction of the relative airflow is different in each diagram.   But not the AoA.  Nish maybe my answer may help, sorry it grew so long.

Part of the problem here is that the title and the first 2 sentences appear to be describing 2 different cases.  Are we holding the AoA constant, or the pitch attitude?  If the AoA, see  the first para of my (overly long) answer.  If the pitch attitude, see esp the para starting with "For an extreme case of this scenario, forget about starting from "cruise flight"".  I would better have pointed out this out sooner and waited for the question to be improved before answering-- oh well next time.

To clarify. The angle of attack is the angle between two vectors: the flight direction vector and the vector on the plane of the wing, which is also directed in the direction of motion (in some calculations, the direction of this vector is considered differently, and correction formulas are used). The higher the angle of attack, the higher the lift force. The higher the speed, the smaller the angle of attack must be to support the weight of the airplane.
The picture you attached only considers the case where the airplane speed is so high that a very small angle of attack is needed to support the weight of the airplane. It will be near zero, but it will only become equal to zero when the airplane is vertically off the ground and there is no wind. As for the change in angle of attack during or after the maneuver... You change the direction of the wing vector, but due to inertia it takes some time for the body to change direction, at that point the angle of attack will increase and then decrease only when both vectors are close to each other.
If this coincides with the direction of your question, feel free to ask follow-up related questions!

RE "It will be near zero, but it will only become equal to zero when the airplane is vertically off the ground and there is no wind"-- I find this sentence hard to follow.  Are trying to describe the situation where a/c is parked stationary on level ground and there is no wind?  Some rewording may help.

@quietflyer I took it to mean a dive straight at the ground: "Pointed vertically at the ground". At which point all "lift" is created by drag, not an angle between airflow and the wing.

@Azendale -- ok, but the wind would be irrelevant.  If you are talking about the external meteorological wind.  Assuming we're not complicating things by including shears / gradients, the wind doesn't matter, could be 100 mph and would make no difference.

I'm finally starting to understand where the OP is coming from-- I believe he is noticing that if the a/c started in the situation in top figure, and then instantaneously pitched up to pitch attitude of middle figure, then presumably the change to the direction of Rel Wind shown in middle figure would not be instantaneous. Which is true, the flight path must curve to a new direction for that to happen, and that is not instantaneous. (Maybe just a second or two, but not inst.)
But as to whether we eventually end up back at the original AoA, or higher, or lower (e.g. zero), depends on what control inputs we make and what part of the flight envelope we are in. For example consider the case where the instant pitch up (which obviously temporarily increases AoA) puts the plane so far on the back side of the power curve that rather than climbing, the sink rate goes way up, so the flight path curves downward. (Maybe we've chopped the throttle too at the same time.)
Well, now if we keep that pitch attitude constant, we'll see the AoA increase further as the flight path curves earthward till reaching a steady-state again. (Not a very pleasant one!) So no you can't make general statements that the AoA will always tend to decrease, maybe all the way to zero, if you pitch the nose up abruptly and then hold that pitch attitude. But my other answer covered a case where that could indeed happen.

The diagrams reproduced in your question appear to be intended to illustrate the same, small (but non-zero) angle-of-attack in all 3 cases.
As for your quote--

Then after you increase the angle of attack and fly at that angle of
attack for some amount of time and the relative airflow changes
accordingly.

I think what you may be missing is that "flying at that angle of attack" doesn't imply flying at a constant pitch attitude.  For simplicity imagine some sort of weird loop1 where we make a special point of holding the AoA constant starting from immediately after we've first "pulled G's" and began making the flight path curve.  Since the direction of the velocity vector will constantly be changing throughout the loop, so will the direction of the relative wind, by the same principle that is illustrated in the diagrams in your question.  So your quote above is arguably true. But AoA is constant, not decreasing.
But transitioning from horizontal flight to a constant nose-higher pitch attitude (one that eventually results in a stabilized climb) is a different situation.  If we constrain the problem that way, one could argue that once the pitch attitude is established, we'll see the angle-of-attack decrease due to the changing direction of the flight path during the time interval that the flight path is still curving upwards.2  We could further constrain the problem in such a way that the angle-of-attack could indeed end up at zero, if we are flying with non-symmetrical airfoil that still generates lift at zero degrees AoA.
Exactly what we'd be doing with the elevator during this time is a different question that we can't really answer without knowing a few more constraints.  (To a first approximation we'd expect to need to be moving the stick or yoke forward during the time interval that the flight path is still curving upwards, but there are many complicating effects that cause exceptions.)  From a practical piloting perspective, it generally wouldn't be fair or useful to say that the mere fact that we are transitioning into a steady-state climb is the fundamental reason that the angle-of-attack is decreasing during this maneuver, or that climbing flight intrinsically tends to offer stall protection, or that climbing flight is generally conducted at a lower angle-of-attack than cruising flight.  The decrease in AoA during the period that the flight path is still curving upwards can be viewed as an artifact of the fact that we've arbitrarily chosen to hold the pitch attitude exactly constant at this moment, rather than the AoA.  Yet flying by reference to pitch attitude rather than purely by reference to AoA is indeed something that we actually do in many real-world instances.
For an extreme case of this scenario, forget about starting from "cruise flight" and forget about the initial increase in angle-of-attack in your quote above, and instead simply imagine that we are established in horizontal slow flight deep on the "back side" of the thrust curve or power curve, with the nose very high3, and then we add still more power (maybe we have an afterburner or two?) to accelerate without allowing the pitch attitude to change.  Given sufficient power, and a non-symmetrical airfoil, we certainly could eventually end up in very high-airspeed climb with zero degrees AoA.  The diagrams included in your question are clearly not intended to illustrate this sort of scenario!
And to frame the problem in yet another way, in a sustained, truly vertical climb the wing actually must indeed be at the zero-lift AoA4,5. Which would indeed protect us from stalling, as long as we constrain ourselves to manipulate the controls in such a way as to stay in that sustained vertical climb.  But looking at extreme cases like that widens the discussion beyond what you seem to be asking about in your question.
So, it's complicated.  We have to be very specific about which variables we are constraining when we ask questions like this.  The fundamental reason an airplane stalls is that the pilot moves the control stick too far aft, thus placing the wing at the stall AoA.6
Footnotes--

It's interesting to consider exactly what such a loop would look like and how different the required control inputs would be from a "normal" well-flown loop.  The answer is far from obvious, and in most aerobatic light planes we can't accurately monitor AoA directly.  A good potential new ASE question?

Take care to note the importance of the word choice here-- "curving"-- a "curve" is not a linear climb at constant rate or even a linear climb with a positive rate of increase in airspeed and vertical speed.  In the thought experiment here, the time that the flight path will actually be "curving" upwards as the pitch attitude remains constant (and the AoA is therefore decreasing) will tend to be brief.  And don't overlook the fact that the upward curve of the flight path will typically, but not necessarily, have been preceded by an initial increase in AoA.

A delta-winged aircraft might be would be a good choice for this demonstration, due to their good flight characteristics at "high alpha".

So long as the aircraft is not tailsliding!  Also, ignoring the need to offset any tail force, thrust line angle, etc.

See for example the first set of diagrams in this related ASE answer: Does lift equal weight in a climb?

This simplified argument is not intended to imply that the stick or yoke position that commands the stall angle-of-attack truly remains constant during turning flight or loops-- see for example this related ASE answer: What has happened to make me experience negative G with the control stick FULL AFT near the top of a loop?

Sorry, just got kind of addicted to this one, will lay off on making any more updates.